Distributive+Property+2C-1

Distributive Property
toc This property is mainly used to multiply or divide more than two numbers/expressions at the same time. Here are the key rules to remember when using the distributive property... a( b+ c) = ab + ac math || If you have any number/expression in front of parentheses with an equation inside the parenthesis, you must first multiply the number/expression on the outside with each individual number/expression inside the parentheses. || math 3(3+1) = (3*3)+(3*1) math math 3(4)= 9+3 math 12 = 12 || (x+y)(a+b)=(x+y)a + (x+y)b math math = xa +ya+xb+yb math || If you have 2 sets of parentheses you can use the **FOIL** method which is just a way to distribute numbers and to get rid of parentheses.
 * = **General Rules** ||= **Explanations** ||= **Examples** ||
 * math
 * math

**F** irst ( **x** +y)( **a** +b) = xa **O** uter ( **x** +y)(a+ **b** ) = xb **I** nner (x+ **y** )( **a** +b) = ya **L** ast (x+ **y** )(a+ **b** ) = yb

So: math (x+y)(a+b)=xa+xb+ya+yb math and then you'd combine like terms. In this case there are none, but if there was... || (3+4)(5+7)=

**F** irst ( **3** +4)( **5** +7) = 3*5 **O** uter ( **3** +4)(5+ **7** ) = 3*7 **I** nner (3+ **4** )( **5** +7) = 4*5 **L** ast (3+ **4** )(5+ **7** ) = 4*7

So, (3+4)(5+7)=15+21+20+28 (7)(12) = 84 84 = 84 ||

__The Box Method__
Another method for multiplying set(s) of parenthesis is the **Box Method**. This involves finding the area of imaginary squares to help you multiply sets of parenthesis in a more organized way. Here's a video that demonstrates this method.

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**Here's another example..**.

(x-5) (x-9)



**Explanation**

First you multiply x*x to get the area of the first small square, x². Then, x * –5 gives you the second small square, which is –5x. Next, x* –9 will give you the third small square, –9x. For the last small square, –5 * –9 gives you a positive 45. Now, add up all the small squares and combine any like terms. to get the area of the whole square. You will end up with x² –5x –9x +45. The only like terms are –5x and –9x, which combine to give a –14x. Now, your final solution is x² –14x + 45. x² –5x –9x +45 = x² –14x +45

=Practice Problems:= (Page 16 of Prentice Hall Mathematics Textbook, #50-52)

50. -(2x+y)-2(-x-y) -2x-y-2x+y =-4x

51. x(3-y)+y(x+6) 3x-xy+xy+6x =9x

52. 4(2x+y)-2(2x+y) 8x+4y-4x-2y =4x+2y